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A football team plays in a stadium that has a seating capacity of $30,000$ spectators. With the ticket price set at $\$79$, average attendance has been $21,692$. A market survey indicates that for each $\$4$ the ticket price is lowered, the average attendance increases by $1000$. Assume that the stadium has a fixed cost of $50,000$ and a variable cost of $2$ per ticket sold.

a) Determine the ticket price(s) that the football stadium breaks even.

b) Algebraically determine the price that maximizes profit for the football stadium.

c) What does the ticket price have to be to ensure that the profit is $\$1,000,000$?

I'm having trouble creating equations for this word problem. I'm not sure where I would start to make solve the 3 questions.

amWhy
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Gulnoor
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1 Answers1

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Let the ticket price be $p$ and the number of people be $n$. $n=21692-250p$, so the profit is $p(21692-250p)-50000$ where the $50000$ is the stadium price. Expanding gives $-250^2+21692p-50000$

a) This is just asking to solve $-250p^2+21692p-50000=0$. Using the quadratic formula you get $p=\frac{21692\pm\sqrt{420542864}}{500}$. Rounding to the nearest cent gives $p=\$2.37,\$84.40$.

b) The min/max of a quadratic equation is $-\frac{b}{2a}$, so it is $p=\frac{5423}{125}$. Rounding to the nearest cent gives $p=\$43.38$.

c) The maximum is at $p=\frac{5423}{125}$, so plugging it in gives $\frac{52567858}{125}$, or $\$420542.86$ rounded to the nearest cent. This is less than $\$1000000$, so the answer is never.

Jefferson Ji
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  • for a) if the ticket price was 84.40 that would mean 20,342 people would attend since every 4 up from 79 one thousand less people attend. If 20,342 attended the variable costs would be 40,684 + 50,000 fixed fee. 20342 * 84.4 = 1,716,864.8 total revenue. the costs would bring down the profit to a huge 1,626,180.8. Not 0 which would be breakeven. What am I missing? – Gulnoor Mar 15 '21 at 08:36